MCAT General Chemistry Question 268: Answer and Explanation

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Question: 268

5. The heat of sublimation of water is 46 kJ/mol. If heat is transferred to the sample by the environment at a rate of 0.1 kJ/min, approximately how long will it take to lyophilize 40 cm³ of frozen water (density = 0.91 g/mL) ?

A. 7.7 hours
B. 15.3 hours
C. 77.0 hours
D. 153.0 hours

Correct Answer: B

Explanation:

B 40 cm³ of ice is 36 g or 2 moles of water. The heat required to sublimate this sample is 46 kJ/mol(2 mol) = 92 kJ. If heat is transferred at 0.1 kJ/min, then 920 minutes are required. Dividing 920 min by 60 min/hour gives just over 15 hours. Overall:

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